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This is from an exercise which my friend and my can't fully wrap our heads around.

Given $X\sim Exp(\lambda)$ what is the distribution of $Y = X - \texttt{floor(}{X}\texttt{)}$

Approach: Compute pdf of $\texttt{floor(}{X}\texttt{)}$ like in this question and then use the formula for computing the difference of two random variables $f_y(y) = \int_0^{\infty} f_X(x)*f_{\texttt{floor(}{X}\texttt{)}}(y-x) dx$

Unfortunately, this integral diverges for $f_{\texttt{floor(}{X}\texttt{)}} = \lambda e^{-\lambda(y-x+1)}$ and $f_X(x) = \lambda e^{-\lambda x}$

Thank you for other approaches and ideas!

Elias
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  • The convolution gives the pdf of the sum of two independent variables. $X$ and $\lfloor X \rfloor$ are not independent. Also notice that $\lfloor X \rfloor$ is a discrete random variable. If you want to write a convolution integral, you need to integrate wrt a measure or write the pdf of $\lfloor X \rfloor$ as an impulse train. – Maxim Jun 23 '19 at 18:46

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A possible approach. Use the law of total probability and partition based on the value of $\lfloor X\rfloor$. Indeed for $y\in [0,1)$, $$ \begin{align*} P(Y\leq y)&=\sum_{k=0}^\infty P(X-\lfloor X\rfloor\leq y, \lfloor X\rfloor=k)\tag{0}\\ &=\sum_{k=0}^\infty P(X\leq k+ y, k\leq X<k+1)\\ &=\sum_{k=0}^\infty P(k\leq X\leq k+y)\\ &=\sum_{k=0}^\infty e^{-\lambda k}-e^{-\lambda(k+y)}\\ &=\sum_{k=0}^\infty e^{-\lambda k}(1-e^{-\lambda y})\\ &=\frac{1-e^{-\lambda y}}{1-e^{-\lambda}} \end{align*} $$ where in (0) we used the law of total probability. Differentiating we get that the density is given by $$ f(y)=\frac{\lambda e^{-\lambda y}}{1-e^{-\lambda}} $$ for $y\in [0,1)$ and equal to zero otherwise.

Sri-Amirthan Theivendran
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  • This is the exact result I got from more intuitive approach. Floor(x) is constant on the intervals [n,n+1[ and hence it is easy to find the density. Sum over all possible values for n and receive the same density. Thanks! – Elias Jun 23 '19 at 16:26